flatmap | 易学教程

Implementation of flatMap() for State transition

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问题 Exercise 6.8, Chiusano and Bjarnason, Functional Programming in Scala , p. 87 asks how one might implement flatMap() for the following trait: trait RNG { def nextInt: (Int, RNG) } type Rand[+A] = RNG => (A, RNG) The answer key gives the following solution: def flatMap[A,B](f: Rand[A])(g: A => Rand[B]): Rand[B] = rng => { val (a, r1) = f(rng) g(a)(r1) // We pass the new state along } Stackoverflow provides many answers to flatMap()/monad questions, but none which for me answered my questions

What is the use case for flatMap vs map in kotlin

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问题 in https://try.kotlinlang.org/#/Kotlin%20Koans/Collections/FlatMap/Task.kt it has sample of using flatMap and map seems both are doing the same thing, is there a sample to show the difference of using flatMap and map ? the data type: data class Shop(val name: String, val customers: List<Customer>) data class Customer(val name: String, val city: City, val orders: List<Order>) { override fun toString() = "$name from ${city.name}" } data class Order(val products: List<Product>, val isDelivered:

Swift: Flatten an array of dictionaries to one dictionary

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问题 In Swift, I am trying to flatten an array of dictionaries into one dictionary i.e let arrayOfDictionaries = [["key1": "value1"], ["key2": "value2"], ["key3": "value3", "key4": "value4"]] //the end result will be: flattenedArray = ["key1": "value1", "key2": "value2", "key3": "value3", "key4": "value4"] I have tried using flatmap, but the type of the returned result is [(String, AnyObject)] and not [String, Object] ie let flattenedArray = arrayOfDictionaries.flatMap { $0 } // type is [(String,

In RxJava, how to pass a variable along when chaining observables?

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问题 I am chaining async operations using RxJava, and I'd like to pass some variable downstream: Observable .from(modifications) .flatmap( (data1) -> { return op1(data1); }) ... .flatmap( (data2) -> { // How to access data1 here ? return op2(data2); }) It seems like a common pattern but I couldn't find information about it. 回答1: The advice I got from the Couchbase forum is to use nested observables: Observable .from(modifications) .flatmap( (data1) -> { return op1(data1) ... .flatmap( (data2) -> {

Is there any difference between flatten and flatMap(identity)?

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问题 scala> List(List(1), List(2), List(3), List(4)) res18: List[List[Int]] = List(List(1), List(2), List(3), List(4)) scala> res18.flatten res19: List[Int] = List(1, 2, 3, 4) scala> res18.flatMap(identity) res20: List[Int] = List(1, 2, 3, 4) Is there any difference between these two functions? When is it appropriate to use one over the other? Are there any tradeoffs? 回答1: You can view flatMap(identity) as map(identity).flatten . (Of course it is not implemented that way, since it would take two

PySpark Throwing error Method getnewargs([]) does not exist

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问题 I have a set of files. The path to the files are saved in a file., say "all_files.txt". Using apache spark, I need to do an operation on all the files and club the results. The steps that I want to do are: Create an RDD by reading "all_files.txt" For each line in "all_files.txt" (Each line is a path to some file), read the contents of each of the files into a single RDD Then do an operation all contents This is the code I wrote for the same: def return_contents_from_file (file_name): return

Is flatMap guaranteed to be lazy? [duplicate]

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问题 This question already has answers here : Why filter() after flatMap() is “not completely” lazy in Java streams? (7 answers) Closed 5 months ago . Consider the following code: urls.stream() .flatMap(url -> fetchDataFromInternet(url).stream()) .filter(...) .findFirst() .get(); Will fetchDataFromInternet be called for second url when the first one was enough? I tried with a smaller example and it looks like working as expected. i.e processes data one by one but can this behavior be relied on? If

spark快速大数据分析之读书笔记-flatmap与map的区别

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【推荐】2019 Java 开发者跳槽指南.pdf(吐血整理) >>> 以前总是分不清楚spark中flatmap和map的区别，现在弄明白了，总结分享给大家，先看看flatmap和map的定义。 map()是将函数用于RDD中的每个元素，将返回值构成新的RDD。 flatmap()是将函数应用于RDD中的每个元素，将返回的迭代器的所有内容构成新的RDD 有些拗口，看看例子就明白了。 val rdd = sc.parallelize(List("coffee panda","happy panda","happiest panda party")) 输入 rdd.map(x=>x).collect 结果 res9: Array[String] = Array(coffee panda, happy panda, happiest panda party) 输入 rdd.flatMap(x=>x.split(" ")).collect 结果 res8: Array[String] = Array(coffee, panda, happy, panda, happiest, panda, party) flatMap说明白就是先map然后再flat，再来看个例子 val rdd1 = sc.parallelize(List(1,2,3,3)) scala> rdd1.map(x=>x

Scala why flatMap treats (x => x) different than (identity)

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问题 First, map treats x => x identical than identity List(1,2,3).map(x => x) //res0: List[Int] = List(1, 2, 3) List(1,2,3).map(identity) //res1: List[Int] = List(1, 2, 3) Now let's transform a List[Option[Int]] into List[Int] discarding all None. We can do that by .flatten . But the point of this question is to understand how flatMap treats identity val maybeNums = List(Some(1), None, Some(-2), None, None, Some(33)) // Works OK, result = List[Int] = List(1, -2, 33) maybeNums.flatMap(x => x)

flatten and flatMap in scala

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问题 I'll like to check if I have understood flatten and flatMap functions correctly. 1) Am I correct that flatten works only when a collection constitutes of other collections. Eg flatten would work on following lists //list of lists val l1 = List(List(1,1,2,-1,3,1,-4,5), List("a","b")) //list of a set, list and map val l2 = List(Set(1,2,3), List(4,5,6), Map('a'->"x",'b'->"y")) But flatten will not work on following val l3 = List(1,2,3) val l4 = List(1,2,3,List('a','b')) val s1 = "hello world"